# 13-1 Arithmetic and Geometric Sequences - navimath · PDF file Find the nth term of the geometric sequence when a = -2 and r =4 If we use 4n-1 we will generate a sequence whose common

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• 13-1 Arithmetic and Geometric Sequences

• ARITHMETIC

SEQUENCES

These are sequences where the difference

between successive terms of a sequence

is always the same number. This number

is called the common difference.

• NOTATION

1st term = t₁

2nd term = t₂

3rd term = t₃

4th term = t₄

nth term = tn

Some people use a to describe the terms of a sequence. However some prefer to use t.

• 3, 7, 11, 15, 19 …

Notice in this sequence that if we find the difference between any term and

the term before it we always get 4. 4 is then called the common difference

and is denoted with the letter d.

d = 4

To get to the next term in the sequence we would add 4 so a recursive

formula for this sequence is:

41  nn aa

The first term in the sequence would be a1 which is sometimes just written as a.

a = 3

• 3, 7, 11, 15, 19 …

+4 +4 +4 +4

Each time you want another term in the sequence you’d add d. This would

mean the second term was the first term plus d. The third term is the first

term plus d plus d (added twice). The fourth term is the first term plus d plus

d plus d (added three times). So you can see to get the nth term we’d take the

first term and add d (n - 1) times.

d = 4

 dnttn 1 Try this to get the 5th term.

𝑎1 = 3

  1916341535 t

• Let’s look at a formula for an arithmetic sequence and see what it tells us.

 14 n

Subbing in the set of positive integers we get:

3, 7, 11, 15, 19, … What is the

common

difference? d = 4

you can see what the common

difference will be in the

formula

We can think of this as a

“compensating term”. Without

it the sequence would start at 4

but this gets it started where

we want it.

4n would generate the multiples of 4. With the - 1 on the

end, everything is back one. What would you do if you

wanted the sequence 2, 6, 10, 14, 18, . . .?  24 n

• Find the nth term of the arithmetic sequence when a = 6 and d = -2

If we use -2n we will generate a sequence whose common

difference is -2, but this sequence starts at -2 (put 1 in for n to get

first term to see this). We want ours to start at 6. We then need

the “compensating term”. If we are at -2 but want 6, we’d need to

 82  n

Check it out by putting in the first few positive integers and

verifying that it generates our sequence.

6, 4, 2, 0, -2, . . . Sure enough---it starts at 6 and has a

common difference of -2

• Let’s try something a little trickier. What if we just know a couple of

terms and they aren’t consecutive?

The fourth term is 3 and the 20th term is 35. Find the first

term and both a term generating formula and a recursive

formula for this sequence. How many differences would you add

to get from the 4th term to the 20th

term?

daa 16420 35 3 Solve this for d d = 2

The fourth term is the first term plus 3

common differences. daa 314 3 (2)

31 a We have all the info we need to express these sequences.

We’ll do it on next slide.

35,3 204  aa

• The fourth term is 3 and the 20th term is 35. Find the first term and

both a term generating formula and a recursive formula for this

sequence.

35,3 204  aa d = 2 31 a

 52 n makes the common difference 2

makes the first term - 3 instead of 2

21  nn aaThe recursive formula would be:

Let’s check it out. If we find n = 4 we should get the 4th term and n = 20

should generate the 20th term.

  35424 a

  35520220 a

• The fourth term is 3 and the 20th term is 35. Find the first term and

both a term generating formula and a recursive formula for this

sequence.

35,3 204  aa

𝑠𝑡𝑒𝑝 1: 𝑓𝑖𝑛𝑑 𝑡ℎ𝑒 𝑐𝑜𝑚𝑚𝑜𝑛 𝑑𝑖𝑓𝑓𝑒𝑟𝑛𝑐𝑒

𝑑 = 35 − 3

20 − 4 =

32

16 = 2

𝑆𝑡𝑒𝑝 2: 𝑢𝑠𝑒 𝑒𝑖𝑡ℎ𝑒𝑟 𝑜𝑛𝑒 𝑜𝑓 𝑡ℎ𝑒 𝑔𝑖𝑣𝑒𝑛 𝑡𝑒𝑟𝑚: 𝑎𝑛 = 2 𝑛 − 4 + 3

Simply: 𝑎𝑛 = 2𝑛 − 5

Or use the other term: 𝑎𝑛 = 2 𝑛 − 20 + 35

Simplify: 𝑎𝑛 = 2𝑛 − 5

• Often in applications we will want the sum of a certain number of terms in an

arithmetic sequence.

The story is told of a grade school teacher In the 1700's that wanted to keep

the numbers from 1 to 100. These numbers are an arithmetic sequence with

common difference 1. Carl Friedrich Gauss was in the class and had the

answer in a minute or two (remember no calculators in those days). This is

what he did:

1 + 2 + 3 + 4 + 5 + . . . + 96 + 97 + 98 + 99 + 100

sum is 101

sum is 101

With 100 numbers there are 50 pairs that add up to 101. 50(101) = 5050

• This will always work with an arithmetic sequence. The formula for the

sum of n terms is:

 nn aa n

S  1 2

n is the number of terms so n/2 would be the number of pairs

first term last term

Let’s find the sum of 1 + 3 +5 + . . . + 59 But how many terms are there?

We can write a formula for the sequence and then figure out what term number

59 is.

•  nn aa n

S  1 2

first term last term

Let’s find the sum of 1 + 3 +5 + . . . + 59

 12 nThe common difference is 2 and the first term is one so: Set this equal to 59 to find n. Remember n is the term number.

2n - 1 = 59 n = 30 So there are 30 terms to sum up.

  900591 2

30 30 S

• GEOMETRIC

SEQUENCES

These are sequences where the ratio of

successive terms of a sequence is always

the same number. This number is called

the common ratio.

• Notice in this sequence that if we find the ratio of any

term to the term before it (divide them) we always get 2.

2 is then called the common ratio and is denoted with the

letter r.

r = 2

To get to the next term in the sequence we would

multiply by 2 so a recursive formula for this sequence is:

12  nn aa

1, 2, 4, 8, 16 . . .

𝑻𝒉𝒊𝒔 𝒊𝒔 𝒂 𝒓𝒆𝒄𝒖𝒓𝒔𝒊𝒗𝒆 𝒇𝒐𝒓𝒎𝒖𝒍𝒂.

•  2  2  2  2

Each time you want another term in the sequence you’d multiply by r. This

would mean the second term was the first term times r. The third term is the

first term multiplied by r multiplied by r (r squared). The fourth term is the

first term multiplied by r multiplied by r multiplied by r (r cubed). So you can

see to get the nth term we’d take the first term and multiply r (n - 1) times.

r = 2

1 nn ara

Try this to get the 5th term.

a = 1

  1621 155  

a

1, 2, 4, 8, 16 . . .

𝑻𝒉𝒊𝒔 𝒊𝒔 𝒂 𝒓𝒆𝒄𝒖𝒓𝒔𝒊𝒗𝒆 𝒇𝒐𝒓𝒎𝒖𝒍𝒂.

• Let’s look at a formula for a geometric sequence and see what it tells us.

  132  n

Subbing in the set of positive integers we get:

-2, -6, -18, -54 … What is the

common ratio? r = 3

you can see what the common

ratio will be in the formula This factor gets us started in

the right place. With n = 1 we’d

get -2 for the first term

3n-1 would generate the powers of 3. With the - 2 in front,

the first term would be

-2(30) =- 2. What would you do if you wanted the

sequence -4, -12, -36, -108, . . .?   134  n

• Find the nth term of the geometric sequence when a = -2 and r =4

If we use 4n-1 we will generate a sequence whose common ratio is 4, but

this sequence starts at 1 (put 1 in for n to get first term to see this). We

want ours to start at -2. We then need the “compensating factor”. We

need to multiply by -2.

  142  n Check it out by putting in the first few positive integers and

verifying that it generates our sequence.

-2, -8, -32, -128, . . .

Sure enough---it starts at -2 and has a

common ratio of 4

• Find the 8th term of 0.4, 0.04. 0.004, . . .

1 nn ara

1.0 4.0

04.0 r

To find the common ratio, take any

term and divide it by the term in

front

  11.04.0  nna

  00000004.01.04.0 188  

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